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**SHORT ANSWER TYPE QUESTIONS**

**Q.11.**** A mass of 2kg is suspended with thread AB (Fig). Thread CD of the same type is attached to the other end of 2 kg mass. Lower thread is pulled gradually, harder and harder in the downward directon so as to apply force on AB. Which of the threads will break and why?**

**Ans.** Thread AB will break. Force on CD is equal to the force (f) applied at D downward, but the force on thread AB is equal to the force F along with force due to mass 2 kg downward. so the force on AB is 2kg more than applied force at D.Hence the thread AB will break up.**Q.12. ****In the above given problem if the lower thread is pulled with a jerk, what happens?****Ans.** Thread CD will break up if CD is pulled with the jerk because pull on thread CD is not transmitted from CD to AB by mass 2 kg due to its inertia of rest.**Q.13. ****Two masses of 5 kg and 3 kg are suspended with help of massless inextensible strings as shown in Fig. Calculate T _{1} and T_{2} when whole system is going upwards with acceleration = 2 m s^{2} (use g = 9.8 m s^{–2}).**

**Ans. **As the whole system is going up with acceleration = a= 2ms^{-2}

Tension in a string is equal and opposite in all parts of a string.

Forces on mass m_{1}

Forces on mass m_{2}**Q.14.** **Block A of weight 100 N rests on a frictionless inclined plane of slope angle 30° (Fig.). A flexible cord attached to A passes over a frictionless pulley and is connected to block B of weight W. Find the weight W for which the system is in equilibrium.****Ans. **Main concept used: On balanced condition i.e., no motion then no frictional force or

f = 0**Solution.**

During equilibrium of A or B

mg sin30° = F

For B is at rest W = F = 50N.**Q.15. ****A block of mass M is held against a rough vertical wall by pressing it with a finger. If the coefficient of friction between the block and the wall is µ and the acceleration due to gravity is g, calculate the minimum force required to be applied by the finger to hold the block against the wall ?****Ans. **Let F force is applied by the finger on a body of mass M to hold rest against the wall. Under the balanced condition

F = N

And f = Mg

⇒ μF = Mg

or is the minimum force to hold the block against the wall at rest.**Q.16. ****A 100 kg gun fires a ball of 1kg horizontally from a cliff of height 500m. It falls on the ground at a distance of 400m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 m s ^{–2})**

Height of the cliff, h=500 m

Mass of the ball, m= 1 kg

Mass of the gun, M=100 kg

Range of the ball, R= 400 m

Acceleration due to gravity, g= 10 m s

Let the final velocity of the ball be u and the of the gun be v.

Now we know, time of flight is

And range is given by:

R = ut

So, u=400/10 = 40 m s

Therefore, the velocity is:

The recoil velocity of the gun is 0.4 m s

From figure (b) y = t

F = 1 N toward Y-axis.

And the acceleration of the elevator is, a = 2 m s

Here, let’s think this problem from the elevator reference frame. For the person inside the elevator going up with acceleration 2 m s

And as the coin return back to its original position (because we are in the elevator reference frame), the net displacement (s) is zero.

So we can use the laws of motion to solve the problem.

Here, a' = a + g = 12 m s

So the time of flight of the coin is 3.33 seconds.

**LONG ANSWER TYPE QUESTIONS**

**Q.1. ****There are three forces F _{1}, F_{2} and F_{3} acting on a body, all acting on a point P on the body. The body is found to move with uniform speed.**

(a) Show that the forces are coplanar.

(b) Show that the torque acting on the body about any point due to these three forces is zero.

Let these lines be F

As the body is moving with uniform speed, the acceleration is zero. So the net of all forces is zero.

F

So, F

This means even F

Thus the forces are coplanar.

(b) As the sum of all forces are zero, the net torque in any direction is zero. For example, torque about “O” is

Τ = OP x (F

Τ = 0 as (F

Hence, torque acting about any point is also zero.

Ans.

Smooth case

Rough case

Acceleration a = g sinθ − µg cosθ

Now, we can differentiate velocity to calculate Forces(F).

Hence, the total force is

Segment: ABC

Here, the frictional force is

f = μmg

And the centripetal force is

Therefore,

Segment:DEF

Again, the centripetal force is equal to friction.

Now, the time taken is distance/speed.

For ABC, the distance is

And for DEF we have

Therefore,

= 86.3

Hence the time taken is 86.3 seconds.

(b) Show that Fr = −mω

x = A cos ωt

y = B sin ωt

The equation of an ellipse is:

Substituting x and y, we find that LHS = RHS.

So the trajectory is an ellipse.

(b) To find force, we need to find acceleration.

This can be found by differentiating r two times with respect to t.

And

We know,

F = ma

⇒ F = -mrω

**Q.6. ****A cricket bowler releases the ball in two different ways(a) Giving it only horizontal velocity, and(b) Giving it horizontal velocity and a small downward velocity.The speed v**

For (b) also

is the total energy of the ball when it hits the ground

So the speed would be the same for both (a) and (b).

∴ Net components along X and Y-axis will be zero.

Resolving all forces along X-axis

Resolving all forces along Y-axis

(a) At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane?

(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to α > θ ?

(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?

(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a?

(b) If angle α > θ, the angle of inclination of the plane with horizontal it will slide down (f upward) as θ is the angle of repose. So net force downward

(c) To keep the box either stationary or just move it up with uniform velocity (1 =0) upward f downward)

(d) The force applied f.to move the box upward with acceleration a,

Give the magnitude and direction of the (g = 10 m s

(a) Force on the floor of the helicopter by the crew and passengers.

(b) Action of the rotor of the helicopter on the surrounding air.

(c) Force on the helicopter due to the surrounding air.

F - mg = ma

Where m is the mass of the crew and a is the upwards acceleration.

(b) Here, we have to consider all the weight if the system. The force due to rotor F

(c) This force is just the reaction force of Fr. So the force due to surrounding air is 6.25 x 10

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